What is the strongest condition you can up with for \(\alpha\) such that the field extension over \(F\) is equal to the ring of polynomials over \(F\). Or in other words \(F(\alpha)=F[\alpha]\)?

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestThe strongest I could come up with is that \(\alpha\) generates a group such that \(\langle\alpha\rangle\cong C_n\) where \(C_n\) is the \(n^\text{th}\) cyclic group of arbitrary order. – Ali Caglayan · 2 years, 9 months ago

Log in to reply

– Ali Caglayan · 2 years, 9 months ago

I have gone further and said that \(\alpha\) has to be algebraic over \(F\).Log in to reply